Metamath Proof Explorer


Theorem 2lgsoddprmlem3d

Description: Lemma 4 for 2lgsoddprmlem3 . (Contributed by AV, 20-Jul-2021)

Ref Expression
Assertion 2lgsoddprmlem3d 7 2 1 8 = 2 3

Proof

Step Hyp Ref Expression
1 6cn 6
2 8cn 8
3 0re 0
4 8pos 0 < 8
5 3 4 gtneii 8 0
6 1 2 5 divcan4i 6 8 8 = 6
7 1 2 mulcli 6 8
8 ax-1cn 1
9 4p3e7 4 + 3 = 7
10 9 eqcomi 7 = 4 + 3
11 10 oveq1i 7 2 = 4 + 3 2
12 4cn 4
13 3cn 3
14 12 13 binom2i 4 + 3 2 = 4 2 + 2 4 3 + 3 2
15 sq4e2t8 4 2 = 2 8
16 2cn 2
17 4t2e8 4 2 = 8
18 12 16 17 mulcomli 2 4 = 8
19 18 oveq1i 2 4 3 = 8 3
20 16 12 13 mulassi 2 4 3 = 2 4 3
21 2 13 mulcomi 8 3 = 3 8
22 19 20 21 3eqtr3i 2 4 3 = 3 8
23 15 22 oveq12i 4 2 + 2 4 3 = 2 8 + 3 8
24 16 13 2 adddiri 2 + 3 8 = 2 8 + 3 8
25 3p2e5 3 + 2 = 5
26 13 16 25 addcomli 2 + 3 = 5
27 26 oveq1i 2 + 3 8 = 5 8
28 23 24 27 3eqtr2i 4 2 + 2 4 3 = 5 8
29 sq3 3 2 = 9
30 df-9 9 = 8 + 1
31 29 30 eqtri 3 2 = 8 + 1
32 28 31 oveq12i 4 2 + 2 4 3 + 3 2 = 5 8 + 8 + 1
33 5cn 5
34 33 2 mulcli 5 8
35 34 2 8 addassi 5 8 + 8 + 1 = 5 8 + 8 + 1
36 df-6 6 = 5 + 1
37 36 oveq1i 6 8 = 5 + 1 8
38 33 a1i 8 5
39 id 8 8
40 38 39 adddirp1d 8 5 + 1 8 = 5 8 + 8
41 2 40 ax-mp 5 + 1 8 = 5 8 + 8
42 37 41 eqtri 6 8 = 5 8 + 8
43 42 eqcomi 5 8 + 8 = 6 8
44 43 oveq1i 5 8 + 8 + 1 = 6 8 + 1
45 32 35 44 3eqtr2i 4 2 + 2 4 3 + 3 2 = 6 8 + 1
46 14 45 eqtri 4 + 3 2 = 6 8 + 1
47 11 46 eqtri 7 2 = 6 8 + 1
48 7 8 47 mvrraddi 7 2 1 = 6 8
49 48 oveq1i 7 2 1 8 = 6 8 8
50 3t2e6 3 2 = 6
51 13 16 50 mulcomli 2 3 = 6
52 6 49 51 3eqtr4i 7 2 1 8 = 2 3